Question

A uniform solid right circular cone of base radius $R$ is joined to a uniform solid hemisphere of radius $R$ and of the same density, as shown in the figure. The centre of mass of the composite solid lies at the centre of base of the cone. The height of the cone is $\sqrt{x}R\cdot$
Find $x\cdot$

Answer 1

Formula used:$y_{cm}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$
To find relation between $h\text{and}R$, we are given that centre of mass of system lies at junction of cone and hemisphere, so we will take origin as well at centre of mass
$y_{cm}=0,y_1=\frac{h}{4},y_2=\frac{3R}{8}$
Since both bodies have same mass density
$\therefore m_1=\rho \times \frac{1}{3}\pi R^{2}h,m_2=\rho \times \frac{2}{3}\pi R^3$
Using formula $\therefore y_{cm}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$
$0=\frac{\rho \times \frac{1}{3}\pi R^{2}h\times \frac{h}{4}-\rho \frac{2}{3}\pi R^3\times \frac{3R}{8}}{\rho \times \frac{1}{3}\pi R^{2}h+\rho \times \frac{2}{3}\pi R^2}$
$\frac{h^2}{12}-\frac{R^2}{4}=0$
$h=\sqrt{3}R$
Comparing with $h=\sqrt{x}R,\text{we get}x=3$
Final Answer:3